   Chapter 17, Problem 71E

Chapter
Section
Textbook Problem

# Consider the following reaction at 25.0°C: 2 NO 2 ( g ) ⇌ N 2 O 4 ( g ) The values of ∆H° and ∆S° are −58.03 kJ/mol and −176.6 J/K · mol, respectively. Calculate the value of K at 25.0°C. Assuming ∆H° and ∆S° are temperature independent, estimate the value of K at 100.0°C.

Interpretation Introduction

Interpretation: The reaction corresponding to the formation of N2O4 , the value of ΔH° and that of ΔS° is given. The value of K at the given temperature is to be calculated. Also, the value of K at the given temperature, assuming that ΔH° and ΔS° are temperature independent, is to be estimated.

Concept introduction: Equilibrium constant K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, the free energy change is,

ΔG=0Q=K

ΔG=ΔG°+RTln(K)

The expression for standard Gibbs free energy, ΔG° , is,

ΔG°=ΔH°TΔS°

Explanation

Explanation

Given

The value of ΔH° is 58.03kJ/mol .

The value of ΔS° is 176.6J/Kmol .

Temperature is 25 °C .

The conversion of degree Celsius (°C) into Kelvin (K) is done as,

T(K)=T(°C)+273

Hence,

The conversion of 25°C into Kelvin is,

T(K)=T(°C)+273T(K)=(25+273)K=298K

The conversion of kilo-joule (kJ) into joule (J) is done as,

1kJ=103J

Hence,

The conversion of 58.03kJ into joule is,

58.03kJ=(58.03×103)J=58.03×103J

The reaction that takes place is,

2NO2(g)N2O4(g)

The formula of ΔG° is,

ΔG°=ΔH°TΔS° (1)

And,

ΔG°=RTln(K) (2)

Where,

• ΔH° is the standard enthalpy of reaction.
• ΔG° is the free energy change.
• T is the given temperature.
• ΔS° is the standard enthalpy of reaction.

Substitute the value of ΔG° from equation (2) into equation (1).

ΔH°TΔS°=RTln(K)ln(K)=(ΔH°TΔS°RT) (3)

Where,

• R is the gas constant (8

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